tan alpha + 2 tan 2 alpha + 4 tan 4 alpha + 8 | vedclass

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(D) We use the identity $	an 	heta - \cot 	heta = \frac{\sin^2 	heta - \cos^2 	heta}{\sin 	heta \cos 	heta} = -2 \cot 2 	heta$. Given expressi

Vedclass · Accepted Answer

$\cot \alpha$

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(D) We use the identity $	an 	heta - \cot 	heta = \frac{\sin^2 	heta - \cos^2 	heta}{\sin 	heta \cos 	heta} = -2 \cot 2 	heta$. Given expression: $S = 	an \alpha + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$. Adding and subtracting $\cot \alpha$: $S = (	an \alpha - \cot \alpha) + \cot \alpha + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha$ $S = -2 \cot 2 \alpha + 2 	an 2 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$ $S = -2(\cot 2 \alpha - 	an 2 \alpha) + 4 	an 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$ Since $\cot 2 \alpha - 	an 2 \alpha = 2 \cot 4 \alpha$,we have: $S = -2(2 \cot 4 \alpha) + 4 	an 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$ $S = -4 \cot 4 \alpha + 4 	an 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$ $S = -4(\cot 4 \alpha - 	an 4 \alpha) + 8 \cot 8 \alpha + \cot \alpha$ Since $\cot 4 \alpha - 	an 4 \alpha = 2 \cot 8 \alpha$,we have: $S = -4(2 \cot 8 \alpha) + 8 \cot 8 \alpha + \cot \alpha$ $S = -8 \cot 8 \alpha + 8 \cot 8 \alpha + \cot \alpha = \cot \alpha$.

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

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